\subsection{Trace}
\begin{definition}
	The \textit{trace} of a square matrix \( A \in M_{n,n}(F) \equiv M_n(F) \) is defined by
	\[
		\tr A = \sum_{i=1}^n a_{ii}
	\]
	The trace is a linear form.
\end{definition}
\begin{lemma}
	\( \tr (AB) = \tr (BA) \) for any matrices \( A, B \in M_n(F) \).
\end{lemma}
\begin{proof}
	We have
	\[
		\tr (AB) = \sum_{i=1}^n \sum_{j=1}^n a_{ij} b_{ji} = \sum_{j=1}^n \sum_{i=1}^n b_{ji} a_{ij} = \tr (BA)
	\]
\end{proof}
\begin{corollary}
	Similar matrices have the same trace.
\end{corollary}
\begin{proof}
	\[
		\tr(P^{-1}AP) = \tr (A P^{-1} P) = \tr A
	\]
\end{proof}
\begin{definition}
	If \( \alpha \colon V \to V \) is linear, we can define the trace of \( \alpha \) as
	\[
		\tr \alpha = \tr [\alpha]_B
	\]
	for any basis \( B \).
	This is well-defined by the corollary above.
\end{definition}
\begin{lemma}
	If \( \alpha \colon V \to V \) is linear, \( \alpha^\star \colon V^\star \to V^\star \) satisfies
	\[
		\tr \alpha = \tr \alpha^\star
	\]
\end{lemma}
\begin{proof}
	\[
		\tr \alpha = \tr [\alpha]_B = \tr [\alpha]_B^\transpose = \tr [\alpha^\star]_{B^\star} = \tr \alpha^\star
	\]
\end{proof}

\subsection{Permutations and transpositions}
Recall the following facts about permutations and transpositions.
\( S_n \) is the group of permutations of the set \( \qty{1, \dots, n} \); the group of bijections \( \sigma \colon \qty{1, \dots, n} \to \qty{1, \dots, n} \).
A transposition \( \tau_{k \ell} = (k, \ell) \) is defined by \( k \mapsto \ell, \ell \mapsto k, x \mapsto x \) for \( x \neq k, \ell \).
Any permutation \( \sigma \) can be decomposed as a product of transpositions.
This decomposition is not necessarily unique, but the parity of the number of transpositions is well-defined.
We say that the signature of a permutation, denoted \( \varepsilon \colon S_n \to \qty{-1, 1} \), is \( 1 \) if the decomposition has even parity and \( -1 \) if it has odd parity.
We can then show that \( \varepsilon \) is a homomorphism.

\subsection{Determinant}
\begin{definition}
	Let \( A \in M_n(F) \).
	We define
	\[
		\det A = \sum_{\sigma \in S_n} \varepsilon(\sigma) A_{\sigma(1) 1} \dots A_{\sigma(n) n}
	\]
\end{definition}
\begin{example}
	Let \( n = 2 \).
	Then,
	\[
		A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \implies \det A = a_{11} a_{22} - a_{12} a_{21}
	\]
\end{example}
\begin{lemma}
	If \( A = (a_{ij}) \) is an upper (or lower) triangular matrix (with zeroes on the diagonal), then \( \det A = 0 \).
\end{lemma}
\begin{proof}
	Let \( (a_{ij}) = 0 \) for \( i > j \).
	Then
	\[
		\det A = \sum_{\sigma \in S_n} \varepsilon(\sigma) a_{\sigma(1) 1} \dots a_{\sigma(n) n}
	\]
	For the summand to be nonzero, \( \sigma(j) \leq j \) for all \( j \).
	Thus,
	\[
		\det A = a_{1 1} \dots a_{n n} = 0
	\]
\end{proof}
\begin{lemma}
	Let \( A \in M_n(F) \).
	Then, \( \det A = \det A^\transpose \).
\end{lemma}
\begin{proof}
	\begin{align*}
		\det A & = \sum_{\sigma \in S_n} \varepsilon(\sigma) a_{\sigma(1) 1} \dots a_{\sigma(n) n}      \\
		       & = \sum_{\sigma^{-1} \in S_n} \varepsilon(\sigma) a_{\sigma(1) 1} \dots a_{\sigma(n) n} \\
		       & = \sum_{\sigma \in S_n} \varepsilon(\sigma^{-1}) a_{1 \sigma(1)} \dots a_{n \sigma(n)} \\
		       & = \sum_{\sigma \in S_n} \varepsilon(\sigma) a_{1 \sigma(1)} \dots a_{n \sigma(n)}      \\
		       & = \det A^\transpose
	\end{align*}
\end{proof}

\subsection{Volume forms}
\begin{definition}
	A volume form \( d \) on \( F^n \) is a function \( d \colon \underbrace{F^n \times \dots \times F^n}_{n \text{ times}} \to F \) satisfying
	\begin{enumerate}
		\item \( d \) is multilinear: for all \( i \in \qty{1, \dots, n} \) and for all \( v_1, \dots, v_{i-1}, v_{i+1}, \dots, v_n \in F^n \), the map from \( F^n \) to \( F \) defined by
		      \[
			      v \mapsto (v_1, \dots, v_{i-1}, v, v_{i+1}, \dots, v_n)
		      \]
		      is linear.
		      In other words, this map is an element of \( (F^n)^\star \).
		\item \( d \) is alternating: for \( v_i = v_j \) for some \( i \neq j \), \( d = 0 \).
	\end{enumerate}
	So an alternating multilinear form is a volume form.
	We want to show that, up to multiplication by a scalar, the determinant is the only volume form.
\end{definition}
\begin{lemma}
	The map \( (F^n)^n \to F \) defined by \( (A^{(1)}, \dots, A^{(n)}) \mapsto \det A \) is a volume form.
	This map is the determinant of \( A \), but thought of as acting on the column vectors of \( A \).
\end{lemma}
\begin{proof}
	We first show that this map is multilinear.
	Fix \( \sigma \in S_n \), and consider \( \prod_{i=1}^n a_{\sigma(i) i} \).
	This product contains exactly one term in each column of \( A \).
	Thus, the map \( (A^{(1)}, \dots, A^{(n)}) \mapsto \prod_{i=1}^n a_{\sigma(i) i} \) is multilinear.
	This then clearly implies that the determinant, a sum of such multilinear maps, is itself multilinear.

	Now, we show that the determinant is alternating.
	Let \( k \neq \ell \), and \( A^{(k)} = A^{(\ell}) \).
	Let \( \tau = ( k \ell ) \) be the transposition exchanging \( k \) and \( \ell \).
	Then, for all \( i, j \in \qty{1, \dots, n} \), \( a_{ij} = a_{i \tau(j)} \).
	We can decompose permutations into two disjoint sets: \( S_n = A_n \cup \tau A_n \), where \( A_n \) is the alternating group of order \( n \).
	Now, note that \( \prod_{i=1}^n a_{\sigma(i) i} + \prod_{i=1}^n a_{(\tau \circ \sigma)(i) i} = 0 \).
	So the sum over all \( \sigma \in A_n \) gives zero.
	So the determinant is alternating, and hence a volume form.
\end{proof}
\begin{lemma}
	Let \( d \) be a volume form.
	Then, swapping two entries changes the sign.
\end{lemma}
\begin{proof}
	Take the sum of these two results:
	\begin{align*}
		d(v_1, \dots, v_i, \dots, v_j, \dots, v_n) & + d(v_1, \dots, v_j, \dots, v_i, \dots, v_n)               \\
		                                           & = d(v_1, \dots, v_i, \dots, v_j, \dots, v_n)               \\
		                                           & + d(v_1, \dots, v_j, \dots, v_i, \dots, v_n)               \\
		                                           & + d(v_1, \dots, v_i, \dots, v_i, \dots, v_n)               \\
		                                           & + d(v_1, \dots, v_j, \dots, v_j, v_n)                      \\
		                                           & = 2 d(v_1, \dots, v_i + v_j, \dots, v_i + v_j, \dots, v_n) \\
		                                           & = 0
	\end{align*}
	as required.
\end{proof}
\begin{corollary}
	If \( \sigma \in S_n \) and \( d \) is a volume form, \( d(v_{\sigma(1)}, \dots, v_{\sigma(n)}) = \varepsilon(\sigma) d(v_1, \dots, v_n) \).
\end{corollary}
\begin{proof}
	We can decompose \( \sigma \) as a product of transpositions \( \prod_{i=1}^{n_\sigma} e_i \).
\end{proof}
\begin{theorem}
	Let \( d \) be a volume form on \( F^n \).
	Let \( A \) be a matrix whose columns are \( A^{(i)} \).
	Then
	\[
		d(A^{(1)}, \dots, A^{(n)}) = \det A \cdot d(e_1, \dots, e_n)
	\]
	So there is a unique volume form up to a constant multiple.
	We can then see that \( \det A \) is the only volume form such that \( d(e_1, \dots, e_n) = 1 \).
\end{theorem}
\begin{proof}
	\[
		d(A^{(1)}, \dots, A^{(n)}) = d\qty(\sum_{i=1}^n a_{i1} e_i, A^{(2)}, \dots, A^{(n)})
	\]
	Since \( d \) is multilinear,
	\[
		d(A^{(1)}, \dots, A^{(n)}) = \sum_{i=1}^n a_{i1} d\qty(e_i, A^{(2)}, \dots, A^{(n)})
	\]
	Inductively on all columns,
	\[
		d(A^{(1)}, \dots, A^{(n)}) = \sum_{i=1}^n \sum_{j=1}^n a_{i1} a_{j2} d\qty(e_i, e_j, A^{(3)}, \dots, A^{(n)}) = \dots = \sum_{1 \leq i_1, \leq \dots \leq n} \prod_{k=1}^n a_{i_\ell k} d(e_{i_1}, \dots e_{i_n})
	\]
	Since \( d \) is alternating, we know that for \( d(e_{i_1}, \dots, e_{i_n}) \) to be nonzero, the \( i_k \) must be different, so this corresponds to a permutation \( \sigma \in S_n \).
	\[
		d(A^{(1)}, \dots, A^{(n)}) = \sum_{\sigma \in S_n} \prod_{k=1}^n a_{\sigma(k) k} \varepsilon(\sigma) d(e_1, \dots, e_n)
	\]
	which is exactly the determinant up to a constant multiple.
\end{proof}

\subsection{Multiplicative property of determinant}
\begin{lemma}
	Let \( A, B \in M_n(F) \).
	Then \( \det(AB) = \det(A) \det(B) \).
\end{lemma}
\begin{proof}
	Given \( A \), we define the volume form \( d_A \colon (F^n)^n \to F \) by
	\[
		d_A(v_1, \dots, v_n) \mapsto \det(A v_1, \dots, A v_n)
	\]
	\( v_i \mapsto A v_i \) is linear, and the determinant is multilinear, so \( d_A \) is multilinear.
	If \( i \neq j \) and \( v_i = v_j \), then \( \det(\dots, A v_i, \dots, A v_j, \dots) = 0 \) so \( d_A \) is alternating.
	Hence \( d_A \) is a volume form.
	Hence there exists a constant \( C_A \) such that \( d_A(v_1, \dots, v_n) = C_A \det(v_1, \dots, v_n) \).
	We can compute \( C_A \) by considering the basis vectors; \( A e_i = A_i \) where \( A_i \) is the \( i \)th column vector of \( A \).
	Then,
	\[
		C_A = d_A(e_1, \dots, e_n) = \det(Ae_1, \dots, Ae_n) = \det A
	\]
	Hence,
	\[
		\det(AB) = d_A(B) = \det A \det B
	\]
\end{proof}

\subsection{Singular and non-singular matrices}
\begin{definition}
	Let \( A \in M_n(F) \).
	We say that
	\begin{enumerate}
		\item \( A \) is \textit{singular} if \( \det A = 0 \);
		\item \( A \) is \textit{non-singular} if \( \det A \neq 0 \).
	\end{enumerate}
\end{definition}
\begin{lemma}
	If \( A \) is invertible, it is non-singular.
\end{lemma}
\begin{proof}
	If \( A \) is invertible, there exists \( A^{-1} \).
	Then, since the determinant is a homomorphism,
	\[
		\det(A A^{-1}) = \det I = 1
	\]
	Thus \( \det A \det A^{-1} = 1 \) and hence neither of these determinants can be zero.
\end{proof}
\begin{theorem}
	Let \( A \in M_n(F) \).
	The following are equivalent.
	\begin{enumerate}
		\item \( A \) is invertible;
		\item \( A \) is non-singular;
		\item \( r(A) = n \).
	\end{enumerate}
\end{theorem}
\begin{proof}
	We have already shown that (i) implies (ii).
	We have also shown that (i) and (iii) are equivalent by the rank-nullity theorem.
	So it suffices to show that (ii) implies (iii).

	Suppose \( r(A) < n \).
	Then we will show \( A \) is singular.
	We have \( \dim \vecspan(A_1, \dots, A_n) < n \).
	Therefore, since there are \( n \) vectors, \( (A_1, \dots, A_n) \) is not free.
	So there exist scalars \( \lambda_i \) not all zero such that \( \sum_i \lambda_i A_i = 0 \).
	Choose \( j \) such that \( \lambda_j \neq 0 \).
	Then,
	\[
		A_j = -\frac{1}{\lambda_j} \sum_{i \neq j} \lambda_i A_i
	\]
	So we can compute the determinant of \( A \) by
	\[
		\det A = \det(A_1, \dots, -\frac{1}{\lambda_j} \sum_{i \neq j} \lambda_i A_i, \dots, A_n)
	\]
	Since the determinant is alternating and linear in the \( j \)th entry, its value is zero.
	So \( A \) is singular as required.
\end{proof}
\begin{remark}
	The above theorem gives necessary and sufficient conditions for invertibility of a set of \( n \) linear equations with \( n \) unknowns.
\end{remark}

\subsection{Determinants of linear maps}
\begin{lemma}
	Similar matrices have the same determinant.
\end{lemma}
\begin{proof}
	\[
		\det (P^{-1} A P) = \det(P^{-1}) \det A \det P = \det A \det (P^{-1} P) = \det A
	\]
\end{proof}
\begin{definition}
	If \( \alpha \) is an endomorphism, then we define
	\[
		\det \alpha = \det [\alpha]_{B, B}
	\]
	where \( B \) is any basis of the vector space.
	This is well-defined, since this value does not depend on the choice of basis.
\end{definition}
\begin{theorem}
	\( \det \colon L(V,V) \to F \) satisfies the following properties.
	\begin{enumerate}
		\item \( \det I = 1 \);
		\item \( \det (\alpha\beta) = \det\alpha \det\beta \);
		\item \( \det \alpha \neq 0 \) if and only if \( \alpha \) is invertible, and in this case, \( \det(\alpha^{-1}) \det \alpha = 1 \).
	\end{enumerate}
	This is simply a reformulation of the previous theorem for matrices.
	The proof is simple, and relies on the invariance of the determinant under a change of basis.
\end{theorem}

\subsection{Determinant of block-triangular matrices}
\begin{lemma}
	Let \( A \in M_k(F) \), \( B \in M_\ell(F) \), \( C \in M_{k, \ell}(F) \).
	Consider the matrix
	\[
		M = \begin{pmatrix}
			A & C \\
			0 & B
		\end{pmatrix}
	\]
	Then \( \det M = \det A \det B \).
\end{lemma}
\begin{proof}
	Let \( n = k + \ell \), so \( M \in M_n(F) \).
	Let \( M = (m_{ij}) \).
	We must compute
	\[
		\det M = \sum_{\sigma \in S_n} \varepsilon(\sigma) \prod_{i=1}^n m_{\sigma(i) i}
	\]
	Observe that \( m_{\sigma(i) i} = 0 \) if \( i \leq k \) and \( \sigma(i) > k \).
	Then, we need only sum over \( \sigma \in S_n \) such that for all \( j \leq k \), we have \( \sigma(j) \leq k \).
	Thus, for all \( j \in \qty{k+1, \dots, n} \), we have \( \sigma(j) \in \qty{k+1, \dots, n} \).
	We can then uniquely decompose \( \sigma \) into two permutations \( \sigma = \sigma_1 \sigma_2 \), where \( \sigma_1 \) is restricted to \( \qty{1, \dots, k} \) and \( \sigma_2 \) is restricted to \( \qty{k+1, \dots, n} \).
	Hence,
	\begin{align*}
		\det M & = \sum_{\sigma_1 \in S_k} \sum_{\sigma_2 \in S_{n-k}} \varepsilon(\sigma) \prod_{i=1}^n m_{\sigma(i) i}                                                         \\
		       & = \sum_{\sigma_1 \in S_k} \sum_{\sigma_2 \in S_{n-k}} \varepsilon(\sigma_1) \varepsilon(\sigma_2) \prod_{i=1}^k m_{\sigma(i) i} \prod_{i=k+1}^n m_{\sigma(i) i} \\
		       & = \sum_{\sigma_1 \in S_k} \varepsilon(\sigma_1) \prod_{i=1}^k m_{\sigma(i) i} \sum_{\sigma_2 \in S_{n-k}} \varepsilon(\sigma_2) \prod_{i=k+1}^n m_{\sigma(i) i} \\
		       & = \det A \det B
	\end{align*}
\end{proof}
\begin{corollary}
	We need not restrict ourselves to just two blocks, since we can apply the above lemma inductively.
	In particular, this implies that an upper-triangular matrix with diagonal elements \( \lambda_i \) has determinant \( \prod_i \lambda_i \).
\end{corollary}
